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Unformatted text preview: By Joey and Kaitlin Yields Theoretical Yield Is the maximum amount of a given product that can be formed when the limiting reagent is completely consumed Experimental Yield Is the amount of product actually obtained from the chemical reaction Percent Yield Is the actual yield of a product as the percentage of the theoretical yield How to Calculate % Yield= Experimental Yield/Theoretical Yield 1 st determine the limiting reagent (the one that runs out first) The limiting reagent determines how much product is possible to be obtained which is the theoretical yield Example 1determine limiting reagent CH4 + 2 O2 CO2 + 2 H2O If your reaction contains 0.50 moles of CH4 and 0.60 mol O2 which one is the limiting reagent? 0.50 mol CH4 x 1 mol CO2 = 0.50 mol of CO2 1 mol of CH4 0.60 mol O2 x 1 mol CO2 = 0.30 mol CO2 2 mol O2 Example 2Theoretical Yield SiO2 + 4HF SiF4 + 2H2O How many grams of SiF4 can be produced from 78.9 g of SiO2 with an excess of HF? SiO2 = 60.01 g/mol SiF4 = 104.01 g/mol 78.9 g SiO2 x 1 mol SiO2 x 1 mol SiF4 x 104.01 g SiF4 = 137 g SiF4 60.01 g/mol SiO2 1 mol SiO2 1 mol SiF4 Example 3Percent Yield Say that you only managed to produce 105.6 g of SiF4 from the reaction from example 2 what would your percent yield be? 105.6 g SiF4 x 100 = 77.1 % 137 g SiF4 Example 4 2 C2H 6 + 7O2 4CO2 + 6H2O Find the percent yield for this reaction starting with 45 g of C2H 6 and 30 g of O2 when only 18.5 g of CO2 are actually produced. C2H 6 =30.07 g/mol O2 = 32.0 g/mol CO2 = 44.01 g/mol 45.0 g C2H 6 x 1 mol C2H6 x 4 mol CO2 x 44.01 g/mol CO2 = 131.7 g CO2 30.07g/molC2H6 2mol C2H6 1 mol CO2 30.0 g O2 x 1 mol O2 x 4 mol CO2 x 44.01 g/mol CO2 = 23.6 g CO2 32.0 g/molO2 7 mol O2 1 mol CO2 Step One: Limiting Reagent Example 4 2 C2H6 + 7O2 4CO2 + 6H2O Find the percent yield for this reaction starting with 45 g of C2H6 and 30 g of O2 when only 18.5 g of CO2 are actually produced. C2H6 =30.07 g/mol O2 = 32.0 g/mol CO2 = 44.011 g/m Experimental Yield 18.5 g CO2 x 100 = 78.4 % Theoretical Yield 23.6 g CO2 30.0 g O2 x 1 mol O2 x 4 mol CO2 x 44.011 g/mol CO2 = 23.6 g CO2 32.0 g/molO2 7 mol O2 1 mol CO2 Step Two: Theoretical Yield You just found it! 23.6 g of CO2 can be produced from 30 g of O2. Step Three: Percent Yield Example 5 2 C2H 6 + 7O2 4CO2 + 6H2O Find the percent yield for this reaction starting with 18.0 ml of C2H 6 with excess O2 and 4.06 x 102 g of CO2 are actually produced. C2H 6 = 30.07 g/mol C2H 6 (density)= 1.34 x 103 g/ml CO2 = 44.01 g/mol 18.0 ml C2H6 x 1.34 x 10 3 g x 1 mol C2H 6 x 4 mol CO2 x 44.011 g CO2 = 7.06 x 102 g ml C2H 6 30.07 g 2 mol C2H 6 1 mol CO2 4.06 x 102 g C2H 6 x 100 = 57.5 % 7.06 x 102 g C2H 6 Example 6 If 6.3 g of H2O is produced in the following Combustion reaction C 3 H 8 + 5O 2 4H 2 O + 3CO 2 containing 14 g of Propane ( C3H 8 ) and 14 g of O2, How many grams will I have left over of O 2 and Propane?...
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This note was uploaded on 02/05/2011 for the course CHEM 151 taught by Professor Staff during the Fall '08 term at Arizona.
 Fall '08
 STAFF
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